The diagonal of a square is changing at a rate of $0.5cm/sec$. Then the rate of change of area, when the area is $400c{m}^2$, is equal to

The correct option is B

$10\sqrt{2}c{m}^2/sec$

Explanation for the correct option:

Finding the rate of change of area:

Let the side of the square be $a$ and its diagonal be $D$.

rate of change of diagonal is given $0.5cm/sec$

$\therefore \frac{dD}{dt}=0.5$

By Pythagoras theorem,

$$\begin{gathered} D^2=a^2+a^2 \\ \Rightarrow D=\sqrt{2a^2} \\ \Rightarrow D=\sqrt{2}a \\ \Rightarrow a=\frac{D}{\sqrt{2}} \end{gathered}$$

now, area of square is $a^2$.

$$\begin{gathered} \Rightarrow Area=a^2 \\ ={\left(\frac{D}{\sqrt{2}}\right)}^2 \\ =\frac{D^2}{2} \end{gathered}$$

Since the area is given as : $400c{m}^2$, equating both we have,

$$\begin{gathered} \therefore \frac{D^2}{2}=400 \\ \Rightarrow D^2=800 \\ \Rightarrow D=\sqrt{800} \\ \Rightarrow D=20\sqrt{2} \end{gathered}$$

So, when the area is $400c{m}^2$ , diagonal is $20\sqrt{2}cm$.

Therefore, rate of change of area is given by:

$$\begin{gathered} \Rightarrow \frac{dA}{dt}=\frac{d}{dt}\left(\frac{D^2}{2}\right) \\ =\frac{1}{2}\times 2D\times \frac{dD}{dt} \\ =20\sqrt{2}\times 0.5 \\ =10\sqrt{2}c{m}^2/sec \end{gathered}$$

Therefore, the correct answer is option (B).